Problem 59 - Precalculus

Question

Weight of a Boat. A force of 630 pounds is needed to pull a speedboat and its trailer up a ramp that has an incline of \(13^{\circ}\) What is the combined weight of the boat and its trailer? (IMAGE CAN'T COPY)

Step-by-Step Solution

Answer

The combined weight of the boat and its trailer is approximately 2800 pounds.

Step 1: Understanding the Problem

To find the combined weight of the boat and its trailer, we need to consider the force needed to pull them up an incline. The force required is given as 630 pounds, and the ramp has an incline of \(13^{\circ}\). This force corresponds to the horizontal component of the boat’s weight when it's on the incline.

Step 2: Setting Up the Equation

The force pulling the speedboat and trailer up the incline is the parallel component of their weight with respect to the incline. The weight of the boat and trailer \(W\) creates a components along the incline which is \(W \sin(\theta)\), where \(\theta\) is the angle of the incline. We set up the equation: \[ W \sin(13^{\circ}) = 630 \text{ pounds} \]

Step 3: Solving for Weight

We need to solve for \(W\). Divide both sides by \(\sin(13^{\circ})\) to isolate \(W\):\[ W = \frac{630}{\sin(13^{\circ})} \]Using a calculator, find \(\sin(13^{\circ}) \approx 0.22495\). Substitute this value into the equation to find \(W\):\[ W = \frac{630}{0.22495} \approx 2800 \text{ pounds} \]

Step 4: Conclusion

The combined weight of the boat and its trailer is approximately 2800 pounds.

Key Concepts

Force Components

When pulling an object up a slope, the force required is not simply the weight of the object. Instead, it is necessary to understand the components of forces that act upon it. These forces can be broken down into specific parts that act in different directions.

For an object on an incline, the gravitational force pulling it downwards can be divided into:

• The parallel component, acting along the incline.
• The perpendicular component, acting perpendicular to the incline.

The parallel component is particularly important because it determines the effort required to move the object uphill. Mathematically, this is represented by the equation: \[ F_{parallel} = W imes ext{sin}( heta) \], where:

• \( F_{parallel} \) is the force parallel to the incline.
• \( W \) is the weight of the object.
• \( \theta \) is the angle of the incline.

Incline Problems

Incline problems involve finding forces acting on objects situated on an inclined plane. The inclination alters the way gravity affects the object, necessitating the breakdown of its weight into components. Understanding these problems is key to identifying how much force is required to move or hold an object in place on the slope.

Let's consider a speedboat and trailer being pulled up an incline of \( 13^{\circ} \). The weight, typically acting vertically, in this case must be projected onto the incline to find the component causing slippage downward.

This specific scenario illustrates the importance of understanding trigonometric angles and how they affect force vectors. The weight of the object along the ramp, its parallel component is calculated using the equation \( W \sin(\theta) \). In a real-world application, this helps determine the actual effort required to move the object uphill.

Weight Calculation

Understanding weight calculation is crucial when analyzing forces on inclines. In many practical situations, like moving a boat uphill, weight is not always directly perceivable and needs to be calculated from observed forces.

In our problem, to find the combined weight of the boat and trailer, we use the relationship between the weight of the object and its parallel component of force on the incline. As established, the formula \( F_{parallel} = W \sin(\theta) \) allows us to determine how much of the weight contributes to the effort of pulling the object.

By rearranging the formula, we can solve for the total weight \( W \) given: \[ W = \frac{F_{parallel}}{\sin(\theta)} \] In this case, with \( F_{parallel} = 630 \) pounds and \( \theta = 13^{\circ} \), we calculate \( W \approx 2800 \) pounds. This illustrates the practical application of trigonometry in calculating complex force interactions and clarifies the real magnitude of the forces involved.